# [LeetCode 173] Binary Search Tree Iterator

### Question

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

### Analysis

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The solution of the iterator applies to a lot of related questions. So make sure you practise this question until you are perfect. You WILL BE ASKED this question at times.

You could read my other post [Question] Iterator of Binary Search Tree.

### Solution

We only need to keep 1 variable in RAM, that is a stack.

### Code

public class BSTIterator {

Stack<TreeNode> stack = new Stack<TreeNode>();

public BSTIterator(TreeNode root) {
while (root != null) {
stack.push(root);
root = root.left;
}
}

/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}

/** @return the next smallest number */
public int next() {
if (!hasNext()) {
return 0;
}
TreeNode next = stack.pop();
TreeNode node = next.right;
while (node != null) {
stack.push(node);
node = node.left;
}
return next.val;
}
}